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Each cereal box contains 1 of 5 uniformly distributed prizes. If you open 10 randomly selected boxes, what is the probability that you will get all 5 prizes?

More generally, is there a formula you can use to solve this when the number of prizes is x and the number of boxes is y?

joriki
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Bart Wisialowski
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  • Hint: In five arbitrary boxes there must be different prices. At the other 5 boxes the kind of prices doesn't matter. – callculus42 Mar 13 '20 at 17:23
  • Inclusion-Exclusion is a good approach. More broadly, this class of problems is referred to as Coupon Collector Problems – lulu Mar 13 '20 at 17:26
  • @JeanMarie I dont think this is negative binomial. Negative binomial applies when there are only two outcomes for each trial and the probability of successful trial is fixed. Here there are 5 outcomes. Or, if you "success" is defined as getting a prize you haven't gotten yet, then the probability of a "success" changes. – Bart Wisialowski Mar 13 '20 at 19:25
  • @lulu Thank you. The link does help but it provides a solution for expected value. I don't t see there how you calculate the probability for the question I posed. It looks like you could use the Markov inequality the article mentions to put bounds on the probability but I cant easily follow that part of the article. – Bart Wisialowski Mar 13 '20 at 19:32
  • Aha, I think I found it here: https://math.stackexchange.com/a/1454749/523382 – Bart Wisialowski Mar 13 '20 at 19:35
  • @Bart Wisialowski, My bad, you are completely right : the random variable "number of bought boxes" it is the sum of geometric laws but with changing parameters (there is a longer mean time to obtain the second prize, then a still longer mean time to get the third, etc. – Jean Marie Mar 13 '20 at 19:46

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